A simple harmonic oscillator has an amplitude $A$ and time period $6 \pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x}$ s, where $x=$__________.
JEE MAIN 2024, Diffcult
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From phasor diagram particle has to move from $\mathrm{P}$ to $Q$ in a circle of radius equal to amplitude of $SHM.$
$ \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} $
$ \phi=\frac{\pi}{6}$
Now, $\frac{\pi}{6}=\omega t$
$\frac{\pi}{6}=\frac{2 \pi}{T} t$
$\frac{\pi}{6}=\frac{2 \pi}{6 \pi} t$
$\mathrm{t}=\frac{\pi}{2}$
So, $\mathrm{x}=2$
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